Time of a falling object when the force of gravity isn't constant

[Shahar]

$$r_0 = 2e^7, r_1 = 1e^7 $$
$$θ_0 = \frac{π}{2}, θ_1 = .000316$$
$$t = -\frac{{r_0}^{3/2}}{\sqrt{2 \ G \ M}} \left [ θ\ -\ sin(θ)cos(θ) \right ]_{π/2}^{.000316}$$
Need to check the math on this part:
$$t \simeq -1999 (.00031 -\ π/2) \simeq 3139 \ seconds $$

I recalculated θ. θ = 0.78
So t = 2528 seconds.

 

[Chestermiller]

I recalculated θ. θ = 0.78
So t = 2528 seconds.

Yes. It's π/4.

Chet

 

[rcgldr]

Using the other substitution:
$$u = \sqrt{\frac{r}{r_0 \ - \ r}} $$
$$r = r_0 - \frac{r_0}{1 + u^2} $$
$$dr = \frac{2 r_0 u \ du}{(1 + u^2)^2}$$
$$t = -\ \frac{ {r_0}^{3/2}}{\sqrt{2 G M}} \int_{u0}^{u1} \frac{2 \ u^{2}}{(1+u^{2})^{2}}du= -\ \frac{ {r_0}^{3/2}}{\sqrt{2 G M}} \left [ \tan^{-1}(u)-\frac{u}{1+u^{2}} \right ] _{u0}^{u1}$$
$$r_0 = 2e7 \ , \ r_1 = 1e7$$
$$u0 = ∞ \ , \ u1 = 1$$
$$tan^{-1}(∞) = π/2 \ , \ ∞ / (1 \ + \ ∞^2) = 0$$
This result now matches the result in post #70:
$$t = -2000 ((π/4 - 1/2) -\ (π/2 - 0)) \simeq 2570.8 \ seconds $$

 

[Shahar]

Using the other substitution:
$$ u = \sqrt{\frac{r}{r_0 \ - \ r}} $$
$$t = -\ \frac{ {r_0}^{3/2}}{\sqrt{2 G M}} \left [ \tan^{-1}(u)-\frac{u}{1+u^{2}} \right ] _{u0}^{u1}$$
r0 = 2e7
r1 = 1e7
u0 = ∞
u1 = 1
$$t \simeq -1999(.2854 - π/2) \simeq 2570$$

This way is much clearer.
I think Ill learn more calculus and then try to get to get to this equation. Or try another simular one( I want to calculate V and t for a free falling object with air resistance
)

 

[rcgldr]

$$ t \simeq 2570.8 $$

I fixed posts #70 and #73.

Or try another simular one( I want to calculate V and t for a free falling object with air resistance)

For this one, the drag related component of acceleration is related to speed^2, not position, so not quite the same. For a vertical drop, the problem can be solved, for the more general case of an object that also has a horizontal component of velocity, you need to use a numerical integration method.

As mentioned before a simpler problem for acceleration based on position is:
acceleration = -x
at time t = 0, x = 0, v = 1 meter / second
You'll be able to solve this and determine x(t), v(t) and a(t).

 

[Chestermiller]

$$ t \simeq 2570 $$

I fixed posts #70 and #73, and now the results are the same, but still not sure since you got 2528.

That's because he used an inaccurate value of 0.78 for π/4. Your answer of 2570 is spot on.

Chet

 

[Shahar]

$$ t \simeq 2570 $$

I fixed posts #70 and #73, and now the results are the same, but still not sure since you got 2528.

$$\frac{{r_0}^{3/2}}{\sqrt{2 G M}} = \frac{{(2e7)}^{3/2}}{\sqrt{2 \times {6.67384e{-11}} \times {1.5e25}}} \simeq 1999$$

For this one, the drag related component of acceleration is related to speed^2, not position, so not quite the same. For a vertical drop, the problem can be solved, for the more general case of an object that also has a horizontal component of velocity, you need to use a numerical integration method.

As mentioned before a simpler problem for acceleration based on position is:
acceleration = -x
at time t = 0, x = 0, v = 1 meter / second
You'll be able to solve this and determine x(t), v(t) and a(t).

I learned a lot from this question and I think,
with a bit more calculus I will be able to find the motion equations with air resistance

 

[rcgldr]

That's because he used an inaccurate value of 0.78 for π/4. Your answer of 2570 is spot on.

I cleaned this up a bit:

Given g = 10 m / s^2 at r = 1e7 m:
g = M G / r^2
G M = r^2 g = (1e7)^2 x 10 = 1e15
2 G M = 2e15
$$\frac{{r_0}^{3/2}}{\sqrt{2 G M}} = \frac{(2e7)^{3/2}}{\sqrt{2e15}} = 2e7 \sqrt{\frac{2e7}{2e15}} = 2e7 \sqrt{\frac{1}{2e8}} = \frac{2e7}{2e4} = 2000$$
$$t = -2000 ((π/4 - 1/2) -\ (π/2 - 0)) = 2000 (π/4 + 1/2) \simeq 2570.7963267949 $$

u = sqrt(r/r0-r) (versus r = r0 sin^2(θ))

This way is much clearer.

I'd recommend going back and looking at the two methods again. Although u = sqrt(r/r0-r) is a typical substitution method, in this case, with a bit of insight, using r = r0 sin^2(θ) replaced the square root component with tan(θ), and integrating 2 sin^2(θ)dθ is simpler than integrating (2 u^2 du) / (1 + u^2)^2 .

 

[PWiz]

So is there no way to express $r$ in terms of $t$ only?

 

[Chestermiller]

So is there no way to express $r$ in terms of $t$ only?

Of course there is. Go to it rcgldr.

Chet

 

[PWiz]

Btw, does this method assume that the planet remains stationary, or is the motion of both bodies included?

 

[PWiz]

I think I've found a neater solution to the problem. If you leave the equation in the form $$t=\int_{r_0}^r \frac{-dr}{\sqrt{\frac{2GM}{r} - \frac{2GM}{r_0}}}$$
and make the substitution $r=\frac{r_0}{sec^2 \theta}$, you get
$$t=r_0 \sqrt{ \frac{r_0}{2GM}} (arccos(\sqrt{\frac{r}{r_0}})+\frac{\sqrt{r} \sqrt{r_0 -r}}{r_0})$$ with no pi or extra -ve signs.
(You can also arrive at this solution by considering the identity $sin(\frac{\pi}{2} - \theta) = cos(\theta)$ . Since the roles of the domain and range are switched with the inverse function, you get the identity $\frac{\pi}{2} - arcsin(\theta) = arccos(\theta)$ and arrive at the solution above.)

And so far, the only progress I've been able to make in using the function $t(r)$ to obtain the function $r(t)$ is the expression the solution in terms of a(n affine?) parameter $\lambda$ :
$$\sqrt{\frac{r}{r_0}}=cos \ \lambda ,$$ and
$$\frac{1}{r_0} \sqrt{\frac{2GM}{r_0}} t = cos \ \lambda (1+ sin \ \lambda)$$
I was thinking about using these Taylor sequences to go further:
$$cos \ \lambda = \sum_{n=0}^∞ \frac{(-1)^n \lambda^{2n}}{(2n)!}$$
$$sin \ \lambda = \sum_{n=0}^∞ \frac{(-1)^n \lambda^{2n+1}}{(2n+1)!}$$
However, this would then mean that the functions $r(t)$, $v(t)$ and $a(t)$ would all have an infinite series, which would look very odd (pardon the parity pun). Is there any other way around this?

On a side note, doesn't the equation $\frac{dr}{dt}=-\frac{dx}{dt}$ assume that $M>>m$ so that $M$ remains stationary relative to $m$?

 

[Shahar]

On a side note, doesn't the equation $\frac{dr}{dt}=-\frac{dx}{dt}$ assume that $M>>m$ so that $M$ remains stationary relative to $m$?

Yes, $M$ remains stationary relative to $m$.
I would be happy to derive an equation for $M$ ~ $m$ so $M$ isn't stationary, but I don't have enough knolige in calculus.

 

[PWiz]

Yes, $M$ remains stationary relative to $m$.
I would be happy to derive an equation for $M$ ~ $m$ so $M$ isn't stationary, but I don't have enough knolige in calculus.

Here, get started :)
http://tutorial.math.lamar.edu/Classes/CalcI/CalcI.aspx

 

[Chestermiller]

I think I've found a neater solution to the problem. If you leave the equation in the form $$t=\int_{r_0}^r \frac{-dr}{\sqrt{\frac{2GM}{r} - \frac{2GM}{r_0}}}$$
and make the substitution $r=\frac{r_0}{sec^2 \theta}$, you get

This is certainly not any "neater" than the substitution I recommended in post #34, $r=r_0sin^2θ$, since, if $r=\frac{r_0}{sec^2 \theta}$, then $r=r_0cos^2θ$, which is essentially the same substitution. Your θ is just π/2 minus my θ. In posts #50, and 66-70, rcgldr employed my substitution, and obtained the same results you have obtained. Your parameter λ is exactly the same as your parameter θ. If you substitute $π/2-θ$ (Chet's θ) for λ in your final equation, you get rcgldr's results.

I was thinking about using these Taylor sequences to go further:
$$cos \ \lambda = \sum_{n=0}^∞ \frac{(-1)^n \lambda^{2n}}{(2n)!}$$
$$sin \ \lambda = \sum_{n=0}^∞ \frac{(-1)^n \lambda^{2n+1}}{(2n+1)!}$$
However, this would then mean that the functions $r(t)$, $v(t)$ and $a(t)$ would all have an infinite series, which would look very odd (pardon the parity pun). Is there any other way around this?

Using a Taylor series expansion is definitely not the way to solve this non-linear algebraic equation for r as a function of t. There are numerous methods available for numerical solution, including the half-interval technique and Newton's method. (Of course, you could say that Newton's method involves expanding to the linear term in a Taylor series, but this is not what I think you had in mind.)

If I wanted r as a function of t, I would just make a table of t(r) vs. r, and plot t as the abscissa and r as the ordinate.

Chet

 

[PWiz]

This is certainly not any "neater" than the substitution I recommended in post #34, $r=r_0sin^2θ$, since, if $r=\frac{r_0}{sec^2 \theta}$, then $r=r_0cos^2θ$, which is essentially the same substitution. Your θ is just π/2 minus my θ. In posts #50, and 66-70, rcgldr employed my substitution, and obtained the same results you have obtained. Your parameter λ is exactly the same as your parameter θ. If you substitute $π/2-θ$ (Chet's θ) for λ in your final equation, you get rcgldr's results.Using a Taylor series expansion is definitely not the way to solve this non-linear algebraic equation for r as a function of t. There are numerous methods available for numerical solution, including the half-interval technique and Newton's method. (Of course, you could say that Newton's method involves expanding to the linear term in a Taylor series, but this is not what I think you had in mind.)

If I wanted r as a function of t, I would just make a table of t(r) vs. r, and plot t as the abscissa and r as the ordinate.

Chet

OK Chet, I guess it was just me trying to avoid pi in the expression. I'd really appreciate if you can give an approximate expression for $r(t)$ though (or is the function implicit?).

 

[Chestermiller]

OK Chet, I guess it was just me trying to avoid pi in the expression. I'd really appreciate if you can give an approximate expression for $r(t)$ though (or is the function implicit?).

The function is implicit. So you really need to solve the non-linear algebraic equation for r(t).

Just make a table with column 1 being r and column 2 being t. Choose values of r in column 1 in equal increments (between the starting value r₀ and the final value at the planet surface). Calculate the corresponding values of t from your equation, and fill them in in column 2. Then plot a graph of r vs t. This is the easiest thing to do. Just be happy that t can be expressed explicitly in terms of r.

Chet